Trả lời đc câu b chưa bạn

nếu rồi cho mình lời giải nha

toán lớp mấy ???

Ta có:\(A=x^2-4x\)

           \(A=x^2-4x+4-4\)

          \(A=\left(x-2\right)^2-4\le-4\)

Dấu = xảy ra khi x - 2 = 0 ; x = 2

   Vậy Min A = - 4 khi x = 2

Ta có:\(B=x^2+x+1\)

           \(B=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

          \(B=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

                   Dấu = xảy ra khi \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Vậy MIn B = 3/4 khi x=-1/2

Ta có:\(C=\left(x+3y-5\right)^2-6xy+26\)

         \(C=x^2+9y^2+25+6xy-10x-30y-6xy+26\)

         \(C=x^2+9y^2-10x-30y+51\)

        \(C=x^2-10x+25+9y^2-30y+25+1\)

         \(C=\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)

Dấu = xảy ra khi \(x-5=0;3y-5=0\Rightarrow x=5;y=\frac{5}{3}\)

                   Vậy Min C = 1 khi x=5;y=5/3

C= (x+3y-5)² - 6xy + 26

   = x² + (3y)² - 25 - 6xy + 26

   = x² - 6xy + (3y)² + (-25+26)

   = x² - 2.x.3y + (3y)² + 1

   = (x-3y)² + 1

Vì (x-3y)² luôn > hoặc = 0

=) (x-3y)² + 1 luôn > hoặc = 1 

vậy GTNN của C là 1 

a. \(2x\left(x-5\right)-x\left(2x+3\right)=26\Rightarrow2x^2-10x-2x^2-3x=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b. \(\left(3y^2-y+1\right)\left(y-1\right)+y^2\left(4-3y\right)=\frac{5}{2}\)

\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3=\frac{5}{2}\)\(\Rightarrow2y=\frac{7}{2}\Rightarrow y=\frac{7}{4}\)

c. \(2x^2+3\left(x+1\right)\left(x-1\right)=5x^2+5x\Rightarrow5x^2-3=5x^2+5x\)

\(\Rightarrow x=-\frac{3}{5}\)

cảm ơn bạn nhiều nhé 

kb vs mình đi 

\(A=\left(x+3y-5\right)^2-6xy+27\)

\(=x^2+9y^2+25+6xy-30y-10x-6xy+27\)

\(=x^2-10x+25+9y^2-30y+25+2\)

\(=\left(x-5\right)^2+\left(3y-5\right)^2+2\)

\(\left(x-5\right)^2\ge0\)

\(\left(3y-5\right)^2\ge0\)

\(\left(x-5\right)^2+\left(3y-5\right)^2+2\ge2\)

\(MinA=2\Leftrightarrow x=5;y=\frac{5}{3}\)

\(A=\left(x+3y-5\right)^2-6xy+27\)

\(=x^2+9y^2+25+6xy-10x-30y-6xy+27\)

\(=\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+2\)

\(=\left(x-5\right)^2+\left(3y-5\right)^2+2\ge2\)

Dấu = khi \(\begin{cases}\left(x-5\right)^2=0\\\left(3y-5\right)^2=0\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}x=5\\y=\frac{5}{3}\end{cases}\)

Vậy Min_A=2 khi \(\begin{cases}x=5\\y=\frac{5}{3}\end{cases}\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+3y-2\right)\)

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

\(=x^2-\left(y+3\right)^2\)

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

\(=\left(3x-y\right)^2-\left(2y\right)^2\)

\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)